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lys-0071 [83]
3 years ago
5

• what is the typical distance between two adjacent pins on a 14-pin dual-in-line ic package?

Physics
1 answer:
muminat3 years ago
7 0

A 14 pin dual-in-line IC package[14 DIL] is an integrated socket which is most popular form of IC package and has a wide range of application in digital electronics.

The 14-pin DIL has two pairs per side and each pair contains seven connecting pins.

The pairs of pins are arranged linearly one after another.The typical dimensions of width is 6.5 mm and the typical dimension of length is 18 mm.

we are asked to calculate the typical distance between two adjacent pins.

The typical distance between two adjacent pins is calculated as-

                                                                 Typical\ distance =\frac{dimensional\ length}{number\ of\ pins\ in\ each\ row}

                                    =\frac{18 mm}{7}

                                    = 2.5714 mm    [ans]                  

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boyakko [2]

Answer:

final displacement = +24484.5 nm

Explanation:

The path difference when 158 bright spots were observed with red light (λ1 = 656.3 nm) is given as;

Δr = 2d2 - 2d1 = 150λ1

So, 2d2 - 2d1 = 150λ1

Dividing both sides by 2 to get;

d2 - d1 = 75λ1 - - - - eq1

Where;

d1 = distance between the fixed mirror and the beam splitter

d2 = position of moveable mirror from splitter when 158 bright spots are observed

Now, the path difference between the two waves when 114 bright spots were observed is;

Δr = 2d'2 - 2d1 = 114λ1

2d'2 - 2d1 = 114λ1

Divide both sides by 2 to get;

d'2 - d1 = 57λ1

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d'2 is the new position of the movable mirror from the splitter

Now, the displacement of the moveable mirror is (d2 - d'2). To get this, we will subtract eq2 from eq1.

(d2 - d1) - (d'2 - d1) = 75λ1 - 57λ2

d2 - d1 - d'2 + d1 = 75λ1 - 57λ2

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We are given;

(λ1 = 656.3 nm) and λ2 = 434.0 nm.

Thus;

d2 - d'2 = 75(656.3) - 57(434)

d2 - d'2 = +24484.5 nm

5 0
3 years ago
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